Question: You have found the following ages (in years) of all 4 zebras at your local zoo: $ 7,\enspace 1,\enspace 9,\enspace 14$ What is the average age of the zebras at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Explanation: Because we have data for all 4 zebras at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $4$ ages and divide by $4$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{4}} x_i}{{4}} $ $ {\mu} = \dfrac{7 + 1 + 9 + 14}{{4}} = {7.8\text{ years old}} $ Find the squared deviations from the mean for each zebra. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $7$ years $-0.8$ years $0.64$ years $^2$ $1$ year $-6.8$ years $46.24$ years $^2$ $9$ years $1.2$ years $1.44$ years $^2$ $14$ years $6.2$ years $38.44$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{0.64} + {46.24} + {1.44} + {38.44}} {{4}} $ $ {\sigma^2} = \dfrac{{86.76}}{{4}} = {21.69\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{21.69\text{ years}^2}} = {4.7\text{ years}} $ The average zebra at the zoo is 7.8 years old. There is a standard deviation of 4.7 years.